To determine compound X, we analyze the details of its reaction with ozone(O3) followed by zinc and water reduction(Zn/H2O). This is an ozonolysis reaction, which breaks down alkenes into carbonyl compounds.
1. Products given are formaldehyde(HCHO) and 2-methyl propanal(CH3CH(CH3)CHO). Formaldehyde indicates one terminal carbon in a double bond, as it is derived from an alkene ending with a -CH=O group.
2. For 2-methyl propanal to form, the remaining carbon chain must include a propyl group attached to a carbon with a methyl group, which was part of a double bond before ozonolysis.
Let's examine the given options:
Compound | Structure | Ozonolysis Products |
---|---|---|
3-Methylbut-1-ene | CH2=C(CH3)CH2CH3 | HCHO, CH3CH(CH3)CHO |
2-Methylbut-1-ene | CH3CH=C(CH3)CH3 | CH3CHO, CH3CH2CHO |
2-Methylbut-2-ene | CH3CH=C(CH3)CH3 | CH3CHO, CH3CHO |
Pent-2-ene | CH3CH=CHCH2CH3 | CH3CHO, CH3CHO |
3. From the table, 3-Methylbut-1-ene yields the required products of formaldehyde and 2-methyl propanal after ozonolysis. Thus, the correct compound X is 3-Methylbut-1-ene.
The reactions which cannot be applied to prepare an alkene by elimination, are
Choose the correct answer from the options given below:
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :
In organic chemistry, an alkene is a hydrocarbon containing a carbon-carbon double bond.[1]
Alkene is often used as synonym of olefin, that is, any hydrocarbon containing one or more double bonds.
Read More: Ozonolysis
Read More: Unsaturated Hydrocarbon